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Atoms, Molecules and Ions

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1 Atoms, Molecules and Ions
Chapter 2 Atoms, Molecules and Ions

2 History of Chemistry Greeks Alchemy

3 Theory vs. Law Theory – human attempts to explain or interpret natural phenomenon Law – summarizes what occurs (observed behavior)

4 Dalton’s Atomic Theory
Elements - made up of atoms Same elements, same atoms. Different elements, different atoms. Chemical reactions involve bonding of atoms

5 Law of Definite Composition
A compound always contains the same proportion of elements by mass

6 Law of Multiple Proportions
Compounds form from specific combinations of atoms H2O vs H2O2

7 Chemical Bonds Holds compounds together Need to be broken for chemical and physical changes to occur

8 Periodic Table Rows (Left to Right) - periods Columns (top to bottom) - groups

9 The Atom Made up of: Protons – (+) charged Electrons – (-) charged
neutrons

10 Periodic Table Atomic Mass Atomic Number Alkali Metals (Group I)
Alkali Earth Metals (Group II) Chalcogens (Group VI) Halogens (Group VII) Noble Gases (Group VIII) Transition Metals

11 Periodic Table Alkaline Metals – Grps. I & II Transition Metals
Non-metals Halogens – Group VII Noble Gases –Group VIII - little chemical activity

12 Periodic Table Lanthanides Actinides

13 Atomic Mass - # at bottom
Periodic Table Atomic Mass - # at bottom how much element weighs Atomic Number - # on top gives # protons = # electrons

14 Periodic Table Atomic Mass
number below the element not whole numbers because the masses are averages of the masses of the different isotopes of the elements

15 Ions Are charged species Result when elements gain electrons or lose electrons

16 2 Types of Ions Anions – (-) charged Cations – (+) charged Example: F-
Example: Na+

17 Highly Important! Gain of electrons makes element (-) = anion Loss of electrons makes element (+) = cation

18 Charges When elements combine, they have to be in the form of IONS.
Cations and anions combine to form compounds. For a neutral compound, the sum of the charges must be ZERO. For a polyatomic ion, the sum of the charges must equal the charge of the ION.

19 Examples In CO2, the charge of C is + 4 In CO, the charge of C is +2.
In KMnO4, since the charge of K is +1, O is -2 so -2 x 4 = -8, Mn must be +7. In (PO4)3-, the charge of O is -2, so -2 x 4= -8, then P must have a charge of +5, so the sum when the charges are added will be -3.

20 Isotopes Are atoms of a given element that differ in the number of neutrons and consequently in atomic mass.

21 Example Isotopes % Abundance 12C % 13C % 14C 11C

22 For example, the mass of C = 12. 01 a. m
For example, the mass of C = a.m.u is the average of the masses of 12C, 13C and 14C.

23 Determination of Aver. Mass
Ave. Mass = [(% Abund./100) (atomic mass)] [(% Abund./100) (atomic mass)]

24 Take Note: If there are more than 2 isotopes, then formula has to be re-adjusted

25 Sample Problem 1 Assume that element Uus is synthesized and that it has the following stable isotopes: 284Uus (283.4 a.m.u.) 34.6 % 285Uus (284.7 a.m.u.) 21.2 % 288Uus (287.8 a.m.u.) %

26 Solution Ave. Mass of Uus = = 97.92 + 60.36 + 127.21
[284Uus] (283.4 a.m.u.)(0.346) [285Uus] +(284.7 a.m.u.)(0.212) [288Uus] +(287.8 a.m.u.)(0.4420) = = a.m.u (FINAL ANS.)

27 Periodic Table Mendeleev – arranged elements in the (.) table

28 Periodic Table Atomic Mass
number below the element not whole numbers because the masses are averages of the masses of the different isotopes of the elements

29 For example, the mass of C = 12. 01 a. m
For example, the mass of C = a.m.u is the average of the masses of 12C, 13C and 14C.

30 Oxidation Numbers Is the charge of the ions (elements in their ion form) Is a form of electron accounting Compounds have total charge of zero (positive charge equals negative charge)

31 Oxidation States Are the partial charges of the ions. Some ions have more than one oxidation states.

32 Oxidation States - generally depend upon the how the element follows the octet rule Octet Rule – rule allowing elements to follow the noble gas configuration

33 Nomenclature - naming of compounds

34 Periodic Table Rows (Left to Right) - periods Columns (top to bottom) - groups

35 Rule 1 – IONIC COMPOUNDS Name metal or first element as is
Metals w/ Fixed Oxidation States Name metal or first element as is - Anion always ends in “–ide”

36 Terminal element or anion
O - oxide P - phosphide N - nitride Se - selenide S - sulfide Cl - chloride F - fluoride I - iodide Br - bromide C - carbide

37 Note Only elements that come directly from the periodic table WILL end in –IDE. POLYATOMIC IONS will be named AS IS.

38 Name the following: CaO - NaCl - MgO - CaS - Na3N -

39 Answers: CaO - calcium oxide NaCl - sodium chloride MgO - magnesium oxide CaS - calcium sulfide Na3N - sodium nitride

40 Where do the subscripts come from?
Answer: From the oxidation states of the ions. Remember: Ions are the species that combine. Target: Compounds! (No charges!)

41 Second Rule II. Ionic Compounds - Metals with no fixed oxidation states (Transition Metals) except for Ag, Zn and Al Metal(Roman #) + 1st syllable + ide Use Roman numerals after the metal to indicate oxidation state

42 Name the following: Copper (I) sulfide Iron (II) oxide Tin (II) iodide
Iron (III) nitride

43 Answers: Copper (I) sulfide Cu2S Iron (II) oxide FeO
Tin (II) iodide SnI2 Iron (III) nitride FeN

44 What about…….? Cesium hydroxide Iron (III) acetate Lithium phosphate
Aluminum Sulfite Lead (II) sulfate Silver nitrate

45 POLYATOMIC IONS Consist of more than 1 element. Have charges.
Ex. SO4 2-, SO3 2-, PO4 3-,PO3 3-

46 Rule 3 – Covalent Compounds
III. For Non-metals (grps IV, V, VI VII), use prefixes. Mono – 1 Hepta - 7 Di Octa - 8 Tri – 3 Nona - 9 Tetra – 4 Deca - 10 Penta – 5 Hexa - 6

47 Rule 3 – Covalent Compounds (only have Non- Metals)
Name 1st element as is. Use prefix, if necessary. Prefix + 1st element + prefix + 1st syllable of anion + ide

48 Name the following compounds
CO2 - carbon dioxide N2O – dinitrogen oxide SO3 – sulfur trioxide N2O5 – dinitrogen pentoxide P2S5 – diphosphorus pentasulfide CO – carbon monoxide

49 Naming Acids Use hydro + 1st syllable + “- ic acid”
I. Acids without Oxygen Use hydro + 1st syllable + “- ic acid” Example: HCl = hydrochloric acid HCN = hydrocyanic acid HBr = hydrobromic acid

50 II. Acids with oxygen Polyatomic “ate” converts to “ic” + acid
Polyatomic “ite” converts to “ous” + acid - H2SO sulfurous acid H2SO4 sulfuric acid HNO3 nitric acid HNO2 nitrous acid H3PO4 phosphoric acid

51 Trick! If anion ends in “ – ate”, acid ends in “ – ic” Example:
HClO4 perchlorate perchloric acid HClO3 chlorate chloric acid

52 Trick! If anion ends in “ – ite”, acid ends in “ – ous” Example:
HClO2 chlorite chlorous acid HClO hypochlorite hypochlorous acid

53 Name the following: HBrO4 (perbromate) HBrO3 (bromate) HBrO2 (bromite) HBrO (hypobromite)

54 Law of Conservation of Mass
Fundamental laws Law of Conservation of Mass Mass is neither created or destroyed Conversion from one form to another

55 Determination of Aver. Mass
Ave. Mass = [(% Abund./100) (atomic mass)] [(% Abund./100) (atomic mass)]

56 Take Note: If there are more than 2 isotopes, then formula has to be re-adjusted

57 Sample Problem 1 Assume that element Uus is synthesized and that it has the following stable isotopes: 284Uus (283.4 a.m.u.) 34.6 % 285Uus (284.7 a.m.u.) 21.2 % 288Uus (287.8 a.m.u.) %

58 Solution Ave. Mass of Uus = = 97.92 + 60.36 + 127.21
[284Uus] (283.4 a.m.u.)(0.346) [285Uus] +(284.7 a.m.u.)(0.212) [288Uus] +(287.8 a.m.u.)(0.4420) = = a.m.u (FINAL ANS.)

59 Chemical Formula Gives the combining whole number ratios of the elements in a compound C6H12O6

60 Structural Formula Gives the spatial arrangement of atoms in the compound Structural formula for H2O is H – O – H

61 Empirical Formula Only gives the types of elements in the compound and the ratio of the elements in the formula

62 Empirical Formula Does not tell exactly how many of the elements are in the compound

63 Molecular Formula Gives you the exact elemental composition of the compound Formula of the compound as it would actually exist.

64 EF vs. MF Sucrose or table sugar: Molecular Formula = C6H12O6
Empirical Formula = CH2O

65 Sample Problem The compound adrenaline contains % C = % H = 6.56 % O = % N = 8.28 by mass. Find the empirical formula.

66 Empirical Formula EF Determination when % Masses are given

67 Steps to Solve for EF Step 1: Sum up all given percentages. If total equals 100%, go to step 2. If total does not equal 100, the missing % is due to one of the component elements. Step 2: Convert Mass % to grams. Step 3: Calculate moles using mole = gram/molar mass

68 Empirical Formula Step 4. To get simplest ratios, divide the moles calculated by the smallest calculated mole. You must have a ration of 1 for at least one of the element. (Follow rule for rounding). Step 5. You now have the ratios or subscripts for the EF.

69 Molecular Formula Detn.
Step 1. Obtain empirical formula mass by adding atomic masses of all elements in empirical formula

70 Molecular Formula Detn.
Step 2. Get ratio by applying the formula below: Molecular Formula = given molar mass Empirical formula mass

71 Molecular Formula Detn.
Step 3. Multiply empirical formula subscripts by obtained ratio

72 Sample Problem Caffeine, a stimulant found in coffee, contains 49.5 % C, 5.15% H, 28.9 % N, and 16.5 % O by mass. The molar mass of the compound is 195 g/mol. Determine the empirical and molecular formula of caffeine.

73 Sample Problem Ibuprofen, a headache remedy, contains % C, 8.80% H, and % O by mass. The molar mass of the compound is 206 g/mol. Determine the empirical and molecular formula of ibuprofen.


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